I am finding a somewhat obscure wording on my functional analysis text. $A$ is a linear operator on a complex Banach space $E$, which I'm almost sure to mean that $A$ is $E\to E$. Then I read that if the operator $R_\lambda=(A-\lambda I)^{-1}$ is defined on all of $E$, then it is bounded. I suspect that the condition that $A$ is bounded is required, because theorem 2 here would -I think- guarantee that. Is that so?
$\infty$ thanks!
Your link doesn't work for me (page unavailable or reached limit for viewing). But, if $D\subset E$ is a subspace, and $A\colon D\to E$ linear such that $A-\lambda I \colon D\to E$ is bijective [that is, $R_\lambda = (A-\lambda I)^{-1}$ is defined on all of $E$], then
For, $R_\lambda$ is a globally defined linear operator on a Banach space, hence by the closed graph theorem, $R_\lambda$ is bounded if and only if it is continuous if and only if its graph is closed. But the graph of $R_\lambda$ is just the flipped graph of $A-\lambda I$, so $R_\lambda$ is bounded if and only if $A-\lambda I$ is closed. But $A-\lambda I$ is closed if and only if $A$ is closed. If $D = E$, then being closed and being bounded (continuous) are the same for $A$.